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my task is to calculate: $$\lim_{n\rightarrow \propto }\sum _{i=1}^{n}\frac{25}{n^{2}}\sqrt{n^{2}-i^{2}}$$I take the constant out, so: $$\lim_{n\rightarrow \propto }\frac{25}{n^{2}}\sum _{i=1}^{n}\sqrt{n^{2}-i^{2}}$$, but I have no idea how to set the boundries for integral, or Am I supposed to do something else?

Thanks for any advice.

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    Looks like a Riemann sum. – Angina Seng Oct 08 '19 at 18:11
  • $\sqrt{n^2-i^2}=n\sqrt{1-\left(\frac{i}{n}\right)^2}$ gives you a Riemann sum. – Thomas Andrews Oct 08 '19 at 18:16
  • @LordSharktheUnknown yes, I know that it is, but I should now set the intervals for integral, so should I put $$\sqrt{n^{2}-i^{2}}$$ =x? –  Oct 08 '19 at 18:17
  • See https://math.stackexchange.com/questions/469885/the-limit-of-a-sum-sum-k-1n-fracnn2k2/469886#469886 – lab bhattacharjee Oct 08 '19 at 18:17
  • @ThomasAndrews, so then $$\lim_{n\rightarrow \propto }\sum {i=1}^{n}\frac{25}{n^{2}}\sqrt{n^{2}-i^{2}}=\lim{n\rightarrow \propto }\sum {i=1}^{n}\frac{25}{n^{2}}n\sqrt{1-(\frac{i}{n})^{2}}=\lim{n\rightarrow \propto }\frac{25}{n}\sum _{i=1}^{n}\sqrt{1-(\frac{i}{n})^{2}}$$? –  Oct 08 '19 at 18:27
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    Yes, so the limit is $$25\int_{0}^{1}\sqrt{1-x^2},dx$$ @Peter And this is the area of a quarter circle of radius $1.$ – Thomas Andrews Oct 08 '19 at 18:40

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